I am sorry but can I ask you for another hint. I understand that what you wrote is true, but what am I supposed to do with it. Using the Caucy Criterion for series, I know that there is an N such that for all n>m>N,
sn-sm< for all epsilon >0. But where do I go from here? Sorry for my slowness in comprehension.

So this is sort of an inductive proof? We show the inequality holds for n=m+1, and since the an terms are non increasing, we know Sn-Sm<epsilon will hold where n>m+1. Correct?

NO! It's not an inductive proof.
We have to show that IF a series [itex]\sum a_n[/itex] converges, then [itex]\lim_{n\rightarrow \infty} a_n= 0[/itex].

So we ASSUME our series is convergent, then we know that for any [itex]\epsilon >0[/itex] we can find an N such that [itex]|Sn-Sm|<\epsilon[/itex] for all n,m>N.

From the above assumption we have to show that:
for any [itex]\epsilon>0[/itex] there exists an N, such that [itex]|a_n|<\epsilon[/itex] whenever n>N.
(This is just the definition of [itex]\lim_{n\rightarrow \infty} a_n= 0[/itex].)

What I meant was. If you plug m=n+1 in |Sm-Sn| and write it in terms of sums, then....